0
\$\begingroup\$

I want to design a current limiter, for both negative and positive voltages, for my thesis project.

My design consists of a voltage source (+/-10V)(X5) and a transimpedance amplifier(X1.)

Using the already existing circuit, in particular using the transimpedance's output, I though that I could use feedback to implement a current limiter.

Circuit

As you can see in the circuit, the transimpedance amplifier's (X1) output is firstly inverted (or buffered depending on the current flowing direction) and then feeds two comparators (X3, X4) which drive the MOSFETs, comparing the inverted (buffered) transimpedance amplifier's output with a reference voltage.

plot

As you can see in the plot, the load's current(red line) is limited as expected at 5 nA (that's because 2.5V across the 500MOhm generates a 5nA current, which is converted in 5V by the transimpedance amplifier.)

As you can see though, the output voltage (green line) is also limited to 2.5V even if the input voltage (blue line) rises to 10V.

I don't expect this since there is an opamp that should maintain the voltage equal to the input due to feedback. I really need that the voltage rises even if the current is limited.

Can you help me understand why this happens? Is this current limiting circuit pooly designed? If yes, can you suggest to me how can I improve it?

EDIT: I know about Ohm's law. But what I want to design is aimed to recreate a usual experiment that scientist do in papers about memristors. They limit the current to prevent the memristor damage, but they increase the electric field, i.e the voltage, on the device.

Memristor Characteristic

That's is what I would like to design and that's why I'm trying to force Ohm's law.

\$\endgroup\$
6
  • 4
    \$\begingroup\$ $$V = I\cdot R$$ I suggest that you study this formula. Kudos for using micro-cap though. On the other hand I think X5 is going to not going to like your method of controlling current. \$\endgroup\$
    – Andy aka
    Commented Jul 15, 2020 at 14:12
  • \$\begingroup\$ This circuit looks like it is designed by an extremely inexperienced person. The current that a certain load takes usually depends on the voltage across that load. So current limiting is achieved by lowering the voltage. You first have to understand and accept that concept before going further. Are you "designing" this just because you need it? Then know that really designing such a circuit/system is often enough for a Thesis by itself. It is not something you can do "just like that". Many have tried and failed. Even an experienced designer would need some days to design this. \$\endgroup\$
    – Bimpelrekkie
    Commented Jul 15, 2020 at 14:14
  • \$\begingroup\$ You're abolutely right. In a lot of paper about memristors though, even if they limit the current upon the memristor, they rises the voltage to increase the electric field in the memristor cell. They even show the graphs about this process. That is why i'm trying to force the Ohm's law in some way. \$\endgroup\$
    – RawCode
    Commented Jul 15, 2020 at 14:31
  • \$\begingroup\$ It is not possible for a power supply to simultaneously control both the current and the voltage that it supplies to a load. Whichever one of those two is controlled by the power supply, the load will determine the other. \$\endgroup\$
    – Solomon Slow
    Commented Jul 15, 2020 at 14:50
  • \$\begingroup\$ @RawCode: The whole point of the memristor is that its resistance goes up when current goes through it. Since the current stays the same (constant current supply) and the resistance goes up, the voltage across the memristor will go up - but only if your memristor really is a memristor. \$\endgroup\$
    – JRE
    Commented Jul 15, 2020 at 15:17

1 Answer 1

Reset to default
1
\$\begingroup\$

It is not possible because of Ohm's Law:

$$V = IR$$

Your load has \$R = 500{\textrm M\Omega}\$

Therefore \$V = I \cdot 500{\textrm M\Omega}\$

If \$I\$ is 5 nA then \$V\$ is 2.5 V, because 5 nA times 500 M\$\Omega\$ is 2.5 V.

Full stop. End of story.

\$\endgroup\$
7
  • \$\begingroup\$ You're abolutely right. In a lot of paper about memristors though, even if they limit the current upon the memristor, they rises the voltage to increase the electric field in the memristor cell. They even show the graphs about this process. That is why i'm trying to force the Ohm's law in some way. \$\endgroup\$
    – RawCode
    Commented Jul 15, 2020 at 14:29
  • \$\begingroup\$ @RawCode (1) Does the current not increase when they increase the voltage? and (2) is your load a memristor? \$\endgroup\$
    – Stack Exchange Supports Israel
    Commented Jul 15, 2020 at 14:32
  • \$\begingroup\$ Yes to both questions, they limit the current to prevent device damage \$\endgroup\$
    – RawCode
    Commented Jul 15, 2020 at 14:33
  • \$\begingroup\$ @RawCode Oops, I meant: Does the voltage not increase when they increase the current? And if your load is a memristor then don't say it's a resistor \$\endgroup\$
    – Stack Exchange Supports Israel
    Commented Jul 15, 2020 at 15:03
  • 1
    \$\begingroup\$ That's the point! They put a MOSFET in series to increase the series resistance, and also limiting the current. They then increase the voltage and the current is limited by the MOSFET which is used as a series resistor. In your opinion, even if they increase the voltage upon the series memristor-MOSFET, the actual voltage drop across the memristor, does actually increase the electric field upon the memristor? \$\endgroup\$
    – RawCode
    Commented Jul 15, 2020 at 15:16

Not the answer you're looking for? Browse other questions tagged or ask your own question.